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Question: A set P contains 2n+1 distinct objects. The number of subsets of P containing not more than n elemen...

A set P contains 2n+1 distinct objects. The number of subsets of P containing not more than n elements is

A

2n

B

22n

C

2n+1

D

2n-1

Answer

22n

Explanation

Solution

Required number of subsets

= 2n+1C0+2n+1C1+2n+1C2+...+2n+1Cn2n + 1C_{0} +^{2n + 1}C_{1} +^{2n + 1}C_{2} + ... +^{2n + 1}C_{n} (To form a

subset of P under the given condition, we have to choose none, one, two,... or at most n elements out of 2n+1)

= 12{22n+1C0+22n+1C1+2 6mu2n+1C2 6mu+....+2 6mu2n+1Cn}\frac{1}{2}\left\{ 2^{2n + 1}C_{0} + 2^{2n + 1}C_{1} + 2\mspace{6mu}^{2n + 1}C_{2}\mspace{6mu} + .... + 2\mspace{6mu}^{2n + 1}C_{n} \right\}

= 12{(22n+1C0+2n+1C2n+1)+ 6mu(2n+1C1 6mu+ 6mu2n+1C2n)+...+(2n+1Cn 6mu+ 6mu2n+1Cn+1)}\frac{1}{2}\left\{ (2^{2n + 1}C_{0} +^{2n + 1}C_{2n + 1}) + \mspace{6mu}(^{2n + 1}C_{1}\mspace{6mu} + \mspace{6mu}^{2n + 1}C_{2n}) + ... + (^{2n + 1}C_{n}\mspace{6mu} + \mspace{6mu}^{2n + 1}C_{n + 1}) \right\}

(∵ 6mumCr= 6mumCm−r)\left( \because\mspace{6mu}^{m}C_{r} = \mspace{6mu}^{m}C_{m - r} \right)

=12(2n+1C0 6mu+ 6mu2n+1C1+2n+1C2+...+2n+1C2n+1)=12 6mux 6mu(22n+1) 6mu= 6mu22n= \frac{1}{2}(^{2n + 1}C_{0}\mspace{6mu} + \mspace{6mu}^{2n + 1}C_{1} +^{2n + 1}C_{2} + ... +^{2n + 1}C_{2n + 1}) = \frac{1}{2}\mspace{6mu} x\mspace{6mu}(2^{2n + 1})\mspace{6mu} = \mspace{6mu} 2^{2n}